A curve in the plane is defined parametrically by the equations $x=t^3-6t$ and $y=\sqrt{2-t^2}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{t}{(3t^2-6)(\sqrt{2-t^2})}$ (Choice B) B $(6-3t^2)(\sqrt{2-t^2})$ (Choice C) C $-\dfrac{t}{\sqrt{2-t^2}}$ (Choice D) D $\dfrac{6-3t^2}{\sqrt{2-t^2}}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=t^3-6t$ and $y=\sqrt{2-t^2}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\sqrt{2-t^2}\right)}{\dfrac{d}{dt}(t^3-6t)} \\\\ &=\dfrac{\left(\dfrac{-\cancel2t}{\cancel2\sqrt{2-t^2}}\right)}{3t^2-6} \\\\ &=-\dfrac{t}{(3t^2-6)(\sqrt{2-t^2})} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{t}{(3t^2-6)(\sqrt{2-t^2})}$.